CHEMISRY (Set-2)

Choose the correct answer :- (1 mark each) 1. NaCl

NaCl crystal structure is–(a) Body centred (b) Face centred (c) Tetragonal (d) Simple cubic

In body centred cubic crystal structure co-ordinate number of atom is- (b) 4 (b) 8 (c) 9 (d) 12

Number of basic types of crystals are- (b) 4 (b) 7 (c) 14 (d) 8

Which of the following increases the rate of physical adsorption- (b) Decrease in temperature (b) Increase in

temperature(c) Decrease in pressure (d) Decrease in surface are

K

The value of velocity constant (K) of reaction depends on- (a) Concentration of Reactants (b) Concentration of products(c) Volume (d) Temperature

BIHAR BOARD CLASS 12 MODEL PAPER

Iron is a– (a) Paramagnetic substance (b) Diamagnetic substance (c) Ferromagnetic substance (d) Ferrimagnetic substance

– / Ea RTK A e – / Ea RTK e/ Ea RTK Ae / Ea RTK e

Arrhenius equation is– (b) – / Ea RTK A e (b) – / Ea RTK e(c) / Ea RTK Ae (d) / Ea RTK e

colligative property

Which of the following is not a collegative property ?(c) Vapour pressure (b) Osmotic pressure(d) Elevation in boiling point (d) Depression in freezing

point

Electrolysis is used in– (b) Electrorefining (b) Electroplating(c) Both (a) and (b) (d) None of these

Colloidal particles of soap in water is- (b) Negatively charged (b) Neutral(c) Positively charged (d) Negatively and positively charged

Milk is– (b) Dispersion of fats in oil (b) Dispersion of fats in

water(c) Dispersion of water in fat (d) Dispersion of water in

oil3CHI

3CHI 3CHI

The antiseptic action of 3CHI is due to7 (b) 3CHI (b) Liberation of free

iodine(c) Iodide ions (d) Iodine and 3CHI both

Which of the following is a secondary alkyl halide ? (b) Isobutyl chloride (b) Isopentyl chloride(c) Neopentyl chloride (d) Isopropyl chloride

2 2CF Cl

2 2CF Cl is used as a/an– (b) Antiseptic (b) Insecticide (c) Analgesic (d)

Refrigerant

Elimination DehydrogenationAddition substitution

The alkyl halide is converted into an alcohol by (b) Elimination (b) Dehydrogenation (c) Addition(d) Substitution

Lucas test is used to distinguish–

(b) Amine (b) Ethers (c) Alcohols(d) Alkylshalides

Fehling test is positive for (b) Acetaldehyde (b) Acetone (c) Ether (d) Amine

Which of the following will not undergo aldol condensation- (b) Acetaldehyde (b) Propanaldehyde(c) Benzaldehyde (d) Trideutero acetaldehyde

2C (H O)x y 2C (H )x y 2(CO) (H )x y 2 2(CO ) (H O)x yThe genral formula of carbohydrate is– (b) 2C (H O)x y (b) 2C (H )x y (c) 2(CO) (H )x y (d)

2 2(CO ) (H O)x y

The number of chiral C- atom in cyclic structure of glucose is- (b) 2 (b) 3 (c) 4 (d) 5

p- s- d- f-Which block of elements are known as transition elements ? (b) p-block (b) s-block (c) d-block (d) f-block

IA IIA IVASodium is a member of which group in periodic table ? (b) Group-IA (b) Group-IIA(c) Group-IVA (d) None of these

4XeF

The shape of 4XeF is– (b) Tetrahedral (b) Square planar (c) Pyramidal (d) LInear

Which one of the following elements is found in free state in nature- (f) Sodium (b) Iron (c) Zince (d) Gold

Which one of the following is an alkaline earth element ? (f) Carbon (b) Calcium (c) Zinc (d) Iron

Which one of the metal is liquid at normal temperature ? (f) Zinc (b) Mercury(g) Sodium (d) Water

2 2HO – CH – CH – OH IUPAC

IUPAC name of 2 2HO – CH – CH – OH is (f) Ethylene glycol (b) Ethane-1, 2-diol(c) Ethyl-1, 2-diol (d) Ethylene diol

Protective sols are– (g) Lyophilic (b) Lyophobic(c) Both (a) and (b) (d) None of these

SOLUTION (1) (b) (2) (b) (3) (b) (4) (a) (5) (d)(6) (c) (7) (a) (8) (a) (9) (c) (10) (a)(11) (b) (12) (b) (13) (d) (14) (d) (15) (d)(16) (c) (17) (a) (18) (c) (19) (a) (20) (c)(21) (c) (22) (a) (23) (b) (24) (d) (25) (b)(26) (b) (27) (b) (28) (a)

Very Short Questions :– (2 marks each)

Q. What do you meant about molecubility of a reaction. Explain with suitableexample.

4 2 2 2NH NO N 2H O1

2 22HI H I

2

2 22NO + O 2NO(2+1=3)

Ans. The no. of reacting species (atoms ions or molecules) taking part in an elementary

reaction, which most collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction.

For example:– 4 2 2 2NH NO N 2H O1

Unimolecular reaction 2 22HI H I

2 Bimolecular reaction

2 22NO + O 2NO(2+1=3)

Trimolecular reaction

Co-ordination number ccp bcc co-ordination

Q. What is meant by the term co-ordination number ? Find the co-ordinationnumber is ccp and bcc.Co-ordination number

Co-ordination number in ccp = 12 Co-ordination number in bcc = 08

Ans. It is defined as the number of nearest neighbours of a particle in a close packed structure.

Co-ordination number in ccp = 12 Co-ordination number in bcc = 08

Q. What is emulsion ? How many types of emulsion.

dispersed phase dispersion medium dispersed dispersed phase dispersion medium dispersed

Ans. Emulsions are colloidal solution of two immicible liquids in which dispersion of tinely divided droplets in another liquid occurs. Emulsion have been classified into two types– (i) Oil in water emulsion – In this dispersed phase is oil while the dispersion

medium is water.Example – Milk (liquid fat is dispersed in water)

(ii) Water in oil Emulsion – In this dispersed phase is water dispersion medium is oil. Example– Butter (water is dispersed into oil)

3 2 |CH CH CHCOOHCl

31 4 10 –[ . ,bK

1 86 . K.Kg/mole]fKQ. Calculate depression in the freezing point of water when 10 gm of

3 2 |CH CH CHCOOHCl

is added to 250 gm of water.

31 4 10 –[ . ,bK 1 86 . K.Kg/mole]fK

22 1

1000= ff

T WT m W W2 = 10 , W1 = 250 , m2 = 122.5 , Kf = 1.86 . / 1000 1.86 10= 250 122.5

fT = 0.607°C fT Ans. We know that 2

2 11000=

ffT WT m W

W2 = 10 gm, W1 = 250 gm, m2 = 122.5 gm, Kf = 1.86 K.Kg/mole 1000 1.86 10= 250 122.5

fT = 0.607°C fT

Q. Write name of two orres of copper.

2CuFeS3 22CuCO Cu(OH)

Ans. The name of two ores of copper:– (a) Copper Pyrites – 2CuFeS (b) Azurite – 3 22CuCO Cu(OH)

Q. What is meant by peptization.

3+ 3+3 3Fe(OH) + Fe [Fe(OH) ]FeAns. The process of converting a freshly prepared precipitate into colloidal form by the

addition of a suitable electrolyte in small amount is called peptization. Peptization involves the adsorption of suitable ions from the electrolyte by the particles of precipitate.

3+ 3+3 3Fe(OH) + Fe [Fe(OH) ]Fe

3NCl 3PCl 5PCl Q. Nitrogen forms only NCl3 but phosphorus forms PCl3 and PCl5 both why ?

d-d-

NCl3 PCl3 PCl5Ans. There is no vacant d-orbital in tha outermost orbit of Nitrogen. Thus nitrogen show

valency only three. There are valent d-orbitals in the outer most orbit of phosphorus and hence it shows variable covalence 3 and 5 in ground state and excited state respectively. Hence nitrogen forms only NCl3 but phosphorus forms PCl3 and PCl5 both.

IUPAC Write down the IUPAC name of the following compounds.

(a) 3| |

3|

CH HH – C C – CH – CH

Br(b)

33| |2|

CHCHH – C C – CH Br

Br

Ans. (a) Trans-4-bromopent-2-ene (b) Cis-1-bromo-2-methylbut-2-ene

Q. Write down the structural formula of the following. (a) 2-Chloro-3-methylpentane (b) 2-Methyl butanoic acid Ans. (a) 3 2 3| |

3CH – C H – C H – CH CH

Cl CH

(b) ||3 2 |

3

OCH – CH – C H – C – OH

CH

Q. How can acetic acid be converted into methyl amine ?Ans. 3 || ||NH H3 3 4 3 2Acetic acid Ammonium acetate

O OCH COOH CH – C – ONH CH – C – NH

2KOH(alc) 3 2Br Methyl amine

CH NH

Q. How are vitamins classified ? Name the vitamin responsible for coagulation of blood.

B CA, D, E, K

E Ans. Vitamins are classified into two groups depending upon their solubility in water fat.

(i) Water soluble vitamins – Vitamin B complex and vitamin C complex.

(ii) FAt soluble vitamins – Vitamin A, D, E, K etc. Vitamin-E responsible for coagulation of blood.

Long Questions :–

Q. State and explain Faraday’s laws of electrolysis.

c tk gm

W Q ( Q ) W ct or, W zct Q ct

z c = 1 t = 1 W = z

1 2,W W 1 2,E E

1 1W E 2 2W E 1 1

2 2W E

W E 2W ct 1 1 W z ct 2 2W z ct 1W 2W

1 12 2

z ct Ez ct E

1 12 2z E

z E z E

Ans. First law of electrolysis:– During electrolysis the deposited mass on the electrode is directly proportional to the

quantity of electricity passing through it. Let W gm of mass is deposited at the electrode after passing camp of current

in t second. Hence, from 1st law of electrolysis. W Q (where Q is the quantity of electricity) W ct or, W zct As we know that Q ct Where z is profitionality constant which is called electrochemical equivalent. If c = 1 amp, t = 1 sec, then W = 2 If 1 amp of current is passed through a solution in one second then the deposited mass

of the substance on the electrode is equal to its electrochemical equivalent. (ii) Second law of electrolysis :– If the same quantity of electricity is passed through the different electrolytic cells connected in a seris then the deposited masses on the electrodes are directly proportional to their chemical equivalents.

Let W1 and W2 be the masses of deposited substances on the electrodes and their chemical equivalents are E1 and E2 respectively then according to

Faraday’s second law 1 1 2 2 W E and W E1 12 2W E

W E ... (i) From 1st law 2W ct

1 1 2 2, W z ct W z ct On putting the value of 1W and 2W in equation (i)

1 12 2

z ct Ez ct E1 12 2z E

z E Thus z E Hence on passing same current through various electrolytes connected in series then, electrochemical equivalent is proportional to their equivalence weights.

Q. Describe principle of production of amonia gas by Haber’s process.Fe/Mo2 2 3450–550N ( ) +3H ( ) 2NH ( )g g g

N2 H2 NH3 2 2 3N ( ) +3H ( ) 2NH ( ) + 24 K.Calg g g

NH3

NH3 N2 H2

°C °C

Ans. Principle behind Haber’s process :– This method involves the direct ambination of Nitrogen and hydrogen as follows

2 2 3N ( ) + 3H ( ) 2NH ( ) + 24 k.calg g g This reaction is reversible, exothermic and followed by decrease in volume.

Accoding to le-chotelier’s principle the optimum condition for greater production of ammonia gas are (i) High pressure – High pressure (200 atm)

shifts the equilibrium. (ii) Low temperature – Since this reaction is

exothermic, the production of ammonaia gas should be high at law temperature. But at low temperature N2 and H2 gases react very slowly. Hence optimum temperature of 450–550°C is mainatained.

(iii) Catalyst – At the optimum temperature 450–550°C the equilibrium may shift to backward direction. To speed up the reaction towards forward direction. Catalyst is used. Finely divided irm + Molybdenum as promoter.

Q. Differentiate between– (a) Mineral and ore (b) Calcium and Roasting (c) Flux and slag

(i) (ii) P, S As

2 2 54P 5O 2P O 2 2S O SO 2 2 34As 3O 2As O

2 2 3MnO SiO MnSiO

2 3SiO CaO CaSiO

2 3SiO + CaO = CaSiO 2 3MnO + SiO = MnSiO

Ans. Difference between mineral and ore:– (a) Mineral – The chemical found in earth crust having high

percentage of any one element is called mineral. Ore – The mineral from which metal can be extracted easily and economically is called ore.

All ores are minerals but all minerals are not ore. (b) Calcination – The processof heating of concentrated ore in absence of air

below m.p is called calcination. In the process of calcination, volatile impurities present in the ore are evaporated out and ore becomes lighter and porous. Roasting – The process of heating of concentrated ore in the reverberatory furnance in persence of air below m.p is called roasting. In the process of roasting– (i) Volatile impurities such as water and organic material are evaporated out. (ii) S, P and As impurities present in the ore are evaporated as oxide

2 2 54P 5O 2P O 2 2S O SO 2 2 34As 3O 2As O

(c) Flux – the foreign substance added in the roasted ore to remove infusible impurities present in the ore is called flux.

Choice of the flux– (i) For basic infusible impurities, acidic flux (SiO2) is used

2 2 3MnO SiO MnSiO (ii) For acidic infusible material, basic flux is used

2 3SiO CaO CaSiO Slag – The fusible material formed due to reaction between gange and flux is called slag. Gang + Flux = Slag 2 3SiO + CaO = CaSiO

1°, 2° 3°Q. What are alcohols 1°, 2°, 3° alcohols ? How will you distinguish them by victor

mayer’s method.1° 1° –OH

3 2 3CH – CH – OH, CH OH

2° 2° –OH

3|3 |

3

CHCH – C – OH

CH

3° 3° –OH

3|3 |

3

CHCH – C – OH

CH

1°, 2° 3°1°

32 AgNOP + I2 2 2 2R–CH OH R – CH R – CH NO 2 2HNO ||R – C – NO

NOH

2° :– 32 AgNOP + I2 2 2 2R –CH – OH R – CHI R – CH – NO

3 2 2AgNO ||

Pseudo Nitrol

R – C – NONO

3° :– 32 2 HNOP + I AgNO3 3 3 2R C – OH R – C – I R C – NO No Reaction

Ans. Primary alcohol (1°) :– It is one in which the –OH group is attached to primary carbon atom.

3 2 3Ethyl alcohol Methyl alcoholCH – CH – OH, CH OH

Secondary alcohol (2°) – Secondary alcohol is that in which –OH group is attached to secondary carbon atom.

|3 |

3Isopropyl alcohol

HCH – C – OH,

CH |

2 5 |3

Butan-2-ol

HC H – C – OH

CH

Tertiary alcohol (3°) – Tertiary alcohol is that in which –OH group is attached to tertiary carbon atom.

3|3 |

3Tertiary Butyl alcohol

CHCH – C – OH

CH

1°, 2° and 3° alcohols can be distinguished by victor Mayer’s method as follows:– 1° alcohol :–

32 AgNOP + I2 2 2 2R–CH OH R – CH R – CH NO 2 2HNO ||

Nitrolic acid

R – C – NONaOH

2° alcohol :–

32 AgNOP + I2 2 2 2R –CH – OH R – CHI R – CH – NO 3 2 2AgNO ||

Pseudo Nitrol

R – C – NONO

3° alcohol :– 32 2 HNOP + I AgNO3 3 3 2R C – OH R – C – I R C – NO No Reduction